\(\dfrac{x-1}{2020}+\dfrac{x-2}{2021}=\dfrac{x-3}{2022}+\dfrac{x-4}{2023}\)

`=>`\(\dfrac{x-1}{2020}+1+\dfrac{x-2}{2021}+1=\dfrac{x-3}{2022}+1\dfrac{x-4}{2023}+1\)

`=>`\(\dfrac{x-1+2020}{2020}+\dfrac{x-2+2021}{2021}=\dfrac{x-3+2022}{2022}+\dfrac{x-4+2023}{2023}\)

`=>`\(\dfrac{x+2019}{2020}+\dfrac{x+2019}{2021}-\dfrac{x+2019}{2022}-\dfrac{x+2019}{2023}=0\)

`=>`\(\left(x+2019\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}\right)=0\)

`=>x+2019=0`

`=>x=-2019`

Vậy `x=-2019`

S = \(\dfrac{4n+5}{2n-3}\)  (n ϵ Z)

S = \(\dfrac{2\left(2n-3\right)+11}{\left(2n-3\right)}\)

S = 2 + \(\dfrac{11}{2n-3}\) 

S nguyên ⇔2n-3 ϵ Ư(11)  = {-11; -1; 1; 11}

⇔ n ϵ { -4; 1; 2; 7}

Ta có \(S=\dfrac{4n+5}{2n-3}=\dfrac{2\left(2n-3\right)+11}{2n-3}=2+\dfrac{11}{2n-3}\)

Để S lầ số nguyên =>\(\dfrac{11}{2n-3}\) nguyên

=> \(11⋮2n-3\) hay 2n-3 \(\in\) Ư(11)

  =>2n-3\(\in\) {1;11;-1;-11}

n\(\in\) {2;7;1;-4}

Lời giải:

a. $0,5.\frac{1}{2}+\frac{3}{4}=\frac{1}{4}+\frac{3}{4}=1$
b.

$\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{27.29}$

$=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{29-27}{27.29}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{29}$

$=1-\frac{1}{29}=\frac{28}{29}$

TL: 

X=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2009}\)

Y=\(\dfrac{2008}{1}\)+\(\dfrac{2007}{2}\)+\(\dfrac{2006}{3}\)+....+\(\dfrac{2}{2007}\)+\(\dfrac{1}{2008}\)

Y=1+ (\(\dfrac{2007}{2}\)+1)+(\(\dfrac{2006}{3}\)+1)+....+(\(\dfrac{1}{2008}\)+1)

Y=\(\dfrac{2009}{2009}\)+\(\dfrac{2009}{2}\)+\(\dfrac{2009}{3}\)+.....+\(\dfrac{2009}{2007}\)+\(\dfrac{2009}{2008}\)

Y=2009(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2008}\)+\(\dfrac{1}{2009}\))

=> \(\dfrac{X}{Y}\)=\(\dfrac{1}{2009}\)

a, Gọi số chẵn là \(2k\) ta có: \(2k.3\) \(⋮\) \(6\)

\(2k.3=6k\); \(6\) \(⋮\) \(6\) \(\Rightarrow6k\) \(⋮\) \(6\) \(\Rightarrow2k.3\) \(⋮\) \(6\)

Vậy 3 số chẵn liên tiếp hay không liên tiếp thì vẫn chia hết cho 6 ( đpcm )

b, Gọi số lẻ là \(2k+1\) ta có: \(3\left(2k+1\right)\) \(⋮̸\)\(6\)

\(3\left(2k+1\right)=6k+3\); \(3\) \(⋮̸\)\(6\) \(\Rightarrow6k+3\) \(⋮̸\)\(6\Rightarrow3\left(2k+1\right)\) \(⋮̸\)\(6\)

Vậy 3 số lẻ liên tiếp hay không liên tiếp thì đều ko chia hết cho 6 ( đpcm )

5. Tính:

\(a)\left[\left(\dfrac{2}{5}\right)^6.\left(\dfrac{2}{5}\right)^5\right]:\left(\dfrac{2}{5}\right)^9\\=\left[\left(\dfrac{2}{5}\right)^{6+5}\right]:\left(\dfrac{2}{5}\right)^9\\ =\left(\dfrac{2}{5}\right)^{11}:\left(\dfrac{2}{5}\right)^9\\ =\left(\dfrac{2}{5}\right)^{11-9}\\ =\left(\dfrac{2}{5}\right)^2\\ =\dfrac{4}{25}\)

\(b)\left[\left(\dfrac{3}{7}\right)^8:\left(\dfrac{3}{7}\right)^7\right].\left(\dfrac{3}{7}\right)\\ =\left[\left(\dfrac{3}{7}\right)^{8-7}\right].\left(\dfrac{3}{7}\right)\\ =\left(\dfrac{3}{7}\right)^1.\left(\dfrac{3}{7}\right)\\ =\left(\dfrac{3}{7}\right)^{1+1}\\ =\left(\dfrac{3}{7}\right)^2\\ =\dfrac{9}{49}\)

\(c)\left[\left(\dfrac{2}{5}\right)^9.\left(\dfrac{2}{5}\right)^4\right]:\left[\left(\dfrac{2}{5}\right)^7.\left(\dfrac{2}{5}\right)^3\right]\\ =\left[\left(\dfrac{2}{5}\right)^{9+4}\right]:\left[\left(\dfrac{2}{5}\right)^{7+3}\right]\\ =\left(\dfrac{2}{5}\right)^{13}:\left(\dfrac{2}{5}\right)^{10}\\ =\left(\dfrac{2}{5}\right)^{13-10}\\ =\left(\dfrac{2}{5}\right)^3\\ =\dfrac{8}{125}.\)

a, \(\left[\left(\dfrac{2}{5}\right)^6.\left(\dfrac{2}{5}\right)^5\right]:\left(\dfrac{2}{5}\right)^9\)

\(=\left(\dfrac{2}{5}\right)^{6+5}:\left(\dfrac{2}{5}\right)^9=\left(\dfrac{2}{5}\right)^{11}:\left(\dfrac{2}{5}\right)^9=\left(\dfrac{2}{5}\right)^{11-9}=\left(\dfrac{2}{5}\right)^2=\dfrac{2^2}{5^2}=\dfrac{4}{25}\)

b, \(\left[\left(\dfrac{3}{7}\right)^8:\left(\dfrac{3}{7}\right)^7\right].\left(\dfrac{3}{7}\right)\)

\(=\left(\dfrac{3}{7}\right)^{8-7}.\dfrac{3}{7}=\dfrac{3}{7}.\dfrac{3}{7}=\dfrac{9}{49}\)

số phần kế hoạch nhà máy cần hoàn thành trong phần cuối là:

     1-( \(\dfrac{4}{15}\)+\(\dfrac{7}{30}\)+\(\dfrac{3}{10}\))= \(\dfrac{1}{5}\)( kế hoạch)